The angle-angle-side criterion for congruent triangles says that: If two angles and a non-included side of one triangle are congruent respectively to two angles and a non-included side of another triangle, then the two triangles are congruent.[br][br]Given triangles [math]\Delta[/math]ABC and [math]\Delta[/math]DEF with [math]\angle[/math]A [math]\cong[/math] [math]\angle[/math]D, [math]\angle[/math]C [math]\cong[/math] [math]\angle[/math]F, and BC [math]\cong[/math] EF. Prove that[br][math]\Delta[/math]ABC and [math]\Delta[/math]DEF.[br][br]Proof:[br]Let [math]\Delta[/math]ABC and [math]\Delta[/math]DEF with [math]\angle[/math]A [math]\cong[/math] [math]\angle[/math]D, [math]\angle[/math]C [math]\cong[/math] [math]\angle[/math]F, and BC [math]\cong[/math] EF. Prove that [math]\Delta[/math]ABC [math]\cong[/math] [math]\Delta[/math]DEF. [br][br]Note: If AC [math]\cong[/math] DF, then the triangles would be congruent because of SAS:[br](AC [math]\cong[/math] DF, BC [math]\cong[/math] EF, [math]\angle[/math]C [math]\cong[/math] [math]\angle[/math]F)[br][br]Case 1: Assume AC < DF. Then by (III-1), there is a point X on DF such that AC [math]\cong[/math] XF. Then[br]ABC [math]\cong[/math] XEF. So [math]\angle[/math]A [math]\cong[/math] [math]\angle[/math]X, but this isn't the case because we assumed [math]\angle[/math]A [math]\cong[/math] [math]\angle[/math]D. [br]Thus, AC is not less than DF. [br][br]Case 2: Assume AC > DF. They by (III-1), there is a point X on AC such that XC [math]\cong[/math] DF. [br]Then XBC [math]\cong[/math] DEF. So [math]\angle[/math]X [math]\cong[/math] [math]\angle[/math]D, but this isn't the case because we assumed [math]\angle[/math]A [math]\cong[/math] [math]\angle[/math]D. Thus, AC is not greater than DF. [br][br]Since AC is neither less than or greater than DF, it must be equal to DF. Therefore, ABC [math]\cong[/math] DEF. [br]